In the figure below,what is the potential dif | vedclass

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(C) In steady state,capacitors act as open circuits,so no current flows through the capacitor branches. The circuit simplifies to a voltage source of

Vedclass · Accepted Answer

${V_{AB}} = 25\,V, {V_{BC}} = 75\,V$

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(C) In steady state,capacitors act as open circuits,so no current flows through the capacitor branches. The circuit simplifies to a voltage source of $100\,V$ connected across the capacitor network.Looking at the circuit,the $3\,\mu F$ and $3\,\mu F$ capacitors are in series,giving an equivalent capacitance of $1.5\,\mu F$. Similarly,the $1\,\mu F$ and $1\,\mu F$ capacitors are in series,giving $0.5\,\mu F$.These two branches are in parallel with each other,and this combination is in series with the $1\,\mu F$ capacitor.However,analyzing the potential divider: The total potential difference across the network is $100\,V$. The branch with $3\,\mu F$ and $3\,\mu F$ (equivalent $1.5\,\mu F$) and the branch with $1\,\mu F$ and $1\,\mu F$ (equivalent $0.5\,\mu F$) are connected between $A$ and $C$ through the $1\,\mu F$ capacitor.By calculating the charge distribution: The potential difference ${V_{AB}}$ is $25\,V$ and ${V_{BC}}$ is $75\,V$.

In the figure below,what is the potential difference between the points $A$ and $B$ and between $B$ and $C$ respectively in steady state?

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